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Why I guess it is true:

I suppose $R$ is a UFD but not PID, then consider coprime ideals $I_1, \ldots,I_n$,
and let $I$ be the product of them. Then the homomorphism $$\phi:R/I \rightarrow R/I_1 \times\dots\times R/I_n$$ such that $$ \phi(a)=(a_1,...,a_n) \space \Longrightarrow \space a-a_k\in I_k$$ it can be proved that phi is a Injection but NOT surjection. This suggests that $$ |R/I|=|R/I_1\times\dots \times R/I_n| \rightarrow \infty $$ so that I guess all quotient rings in $R$ are infinite set. I cannot prove it, I don't even know whether it is correct.

user26857
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Lake Oliver
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  • Hint: consider $R = \Bbb Z[x].,$ Extreme examples are rings whose quotients are all finite - known as called residually finite or FNP rings (FNP = Finite Norm Property). See here for some results on such. – Bill Dubuque Nov 27 '21 at 09:53

1 Answers1

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Welcome to MSE!

I'm not sure what you mean by $|R/I| = |R/I_1 \times \cdots \times R/I_n| \to \infty$, but I do know that your claim is false.

Let $k$ be a finite field. Then $k[x,y]$ is a UFD but not a PID (do you see why?), yet the quotient $k[x,y] / (x,y) \cong k$ is finite (again, do you see why?).

I hope this helps ^_^

Chris Grossack
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